Source: High school physics (Chinese)
Problem Sets:
Problem
A person pushes an object with a mass of $m = 3.0$ kg at a constant velocity for a distance of $s = 10$ m across a horizontal floor. The pushing force is directed downwards at an angle of $\theta = 30^\circ$ relative to the displacement. The coefficient of kinetic friction between the object and the floor is $\mu = 0.20$.
The object moves at a constant velocity, so the net force is zero. Let $F$ be the pushing force. Sum of vertical forces: $N - mg - F \sin\theta = 0 \implies N = mg + F \sin\theta$. Sum of horizontal forces: $F \cos\theta - f = 0 \implies F \cos\theta = f$. The kinetic friction force is $f = \mu N = \mu(mg + F \sin\theta)$. Substituting $f$ into the horizontal force equation:
$$F \cos\theta = \mu(mg + F \sin\theta)$$Solving for $F$:
$$F(\cos\theta - \mu \sin\theta) = \mu mg \implies F = \frac{\mu mg}{\cos\theta - \mu \sin\theta}$$The work done by the person is $W_F = Fs \cos\theta$.
$$W_F = \left( \frac{\mu mg}{\cos\theta - \mu \sin\theta} \right) s \cos\theta = \frac{\mu mgs \cos\theta}{\cos\theta - \mu \sin\theta}$$Substituting the given values ($g = 9.8$ m/s$^2$):
$$W_F = \frac{0.20 \cdot 3.0 \text{ kg} \cdot 9.8 \text{ m/s}^2 \cdot 10 \text{ m} \cdot \cos(30^\circ)}{\cos(30^\circ) - 0.20 \cdot \sin(30^\circ)} = \frac{58.8 \cdot 0.866}{0.866 - 0.20 \cdot 0.5} \approx 66 \text{ J}$$